# Question #85af2

Feb 28, 2017

$= 20 {e}^{0.1 x} + C$

#### Explanation:

In terms of actually understanding, I find it is easier to look at the reverse process first, namely differentiation. We know for example that, for some constant $\alpha$:

$\frac{d}{\mathrm{dx}} \left({e}^{\alpha x}\right) = \alpha {e}^{\alpha x} q \quad \triangle$

(Incidentally, this follows from the more general conclusion that: $\frac{d}{\mathrm{dx}} \left({e}^{f \left(x\right)}\right) = f ' \left(x\right) {e}^{f \left(x\right)}$,..., which you can reach very quickly by using logarithmic differentiation.)

It follows from integrating $\triangle$ that:

$\textcolor{red}{\int \frac{d}{\mathrm{dx}} \left({e}^{\alpha x}\right) \setminus \mathrm{dx}} = \int \alpha {e}^{\alpha x} \setminus \mathrm{dx}$

Or, by realising that in the red term the integration and differentiation "cancel out" (by the Fundamental Theorem of Calculus), and then reversing the order:

$\int \alpha {e}^{\alpha x} \setminus \mathrm{dx} = {e}^{\alpha x} + C$

Or, moving the constant across:
$\int {e}^{\alpha x} \setminus \mathrm{dx} = \frac{1}{\alpha} {e}^{\alpha x} + C$

Then, adding a new constant $\beta$, to reflect your problem:
$\int \beta {e}^{\alpha x} \setminus \mathrm{dx} = \frac{\beta}{\alpha} {e}^{\alpha x} + C$

So with $\alpha = 0.1$, $\beta = 2$:

$\int 2 {e}^{0.1 x} \setminus \mathrm{dx} = \frac{2}{0.1} {e}^{0.1 x} + C$
$= 20 {e}^{0.1 x} + C$

That's all rather mechanical and somewhat dry, but I think it might help find a way of doing these that doesn't involve much memory.